%!TEX program = xelatex
%!TEX TS-program = xelatex
%!TEX encoding = UTF-8 Unicode

\documentclass[12pt,t,aspectratio=169]{beamer}
%Other possible values are: 1610, 149, 54, 43 and 32. By default, it is to 128mm by 96mm(4:3).
%run XeLaTeX to compile.

\input{wang-slides-preamble.tex}

\begin{document}

\title{高等代数一}
\subtitle{5-行最简形、行初等变换、线性方程组的通解}
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
%\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{{\ppr 2022年10月8日} }

\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{内容提要 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}
\item  高斯消元法
\item  线性方程组的系数矩阵和增广矩阵
\item  三类行初等变换
\item  行最简形
\item  非齐次线性方程组导出的齐次线性方程组
\item  基础解系
\item  线性方程组的通解
\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.1. 高斯消元法 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item 例子1：用高斯消元法求解线性方程组
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rrrrr}
x_1 & + x_2 & +2x_3 &=& 4, \\
x_1 &+ 3x_2 &+ 4x_3 &=& 5, \\
2x_1 &+ x_2 &+ x_3 &=& 6. \\
\end{array}\right.
\end{eqnarray*}
}

\item  解答：将第一个方程的 $(-1)$ 倍加到第二个方程，又将第一个方程的 $(-2)$ 倍加到第三个方程，可见保持相同的解集，
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rrrrr}
x_1 & + x_2 & +2x_3 &=& 4, \\
x_1 &+ 3x_2 &+ 4x_3 &=& 5, \\
2x_1 &+ x_2 &+ x_3 &=& 6. \\
\end{array}\right.
\Longleftrightarrow
\left\{\begin{array}{rrrrr}
x_1 & + x_2 & + 2x_3 &=& 4, \\
       &  2x_2 & + 2x_3 &=& 1, \\
       &  - x_2 & -  3x_3 &=& -2. \\
\end{array}\right.
\end{eqnarray*}
}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.2. 系数矩阵与增广矩阵 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  高斯消元法的思路就是通过这些同解变形，化简未知数的系数。

\item  我们看到未知数的位置总是保持不变的，因此只要记录未知数的系数和等式右边的常数的变化就可以了。

\item  这个线性方程组的{\color{red}系数矩阵}与{\color{red}增广矩阵}分别为
{\footnotesize 
\begin{eqnarray*}
A=\begin{bmatrix} 1&1&2 \\ 1&3&4 \\ 2&1&1  \end{bmatrix}, \hspace{0.3cm}
\overline{A} = \begin{bmatrix} 1&1&2&4 \\ 1&3&4&5 \\ 2&1&1&6  \end{bmatrix} = (A,B). 
\end{eqnarray*}
}

\item  {\color{red}高斯消元法}用矩阵来写就是{\color{red}行初等变换}方法。对增广矩阵作行初等变换，先从上往下，再从下往上，目标是为了得到{\color{red}行最简形}。


\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.3. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  线性方程组的两种写法：
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rrrrr}
x_1 & + x_2 & +2x_3 &=& 4, \\
x_1 &+ 3x_2 &+ 4x_3 &=& 5, \\
2x_1 &+ x_2 &+ x_3 &=& 6. \\
\end{array}\right.
\Longleftrightarrow
\begin{bmatrix} 1&1&2 \\ 1&3&4 \\ 2&1&1  \end{bmatrix}
\begin{bmatrix} x_1 \\ x_2 \\ x_3   \end{bmatrix}
=\begin{bmatrix} 4 \\ 5 \\ 6   \end{bmatrix}
\Longleftrightarrow
AX=B.
\end{eqnarray*}
}

\item  三类{\color{red}行初等变换}：
\begin{enumerate}
\item  第一类：交换两行。
\item  第二类：将某一行乘以一个非零常数。
\item  第三类：将某一行乘以一个常数加到另一行。
\end{enumerate}

\item  {\color{red}定理：行初等变换不改变线性方程组的解集。} 


\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.4. 例子1的解答 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

{\footnotesize 
\begin{eqnarray*}
%&& 
\overline{A} = 
\begin{bmatrix} 1&1&2&4 \\ 1&3&4&5 \\ 2&1&1&6  \end{bmatrix}
\xrightarrow[\text{第一行乘以 $-2$ 加到第三行}]{\text{第一行乘以 $-1$ 加到第二行}}
\begin{bmatrix} 1&1&2&4 \\ 0&2&2&1 \\ 0&-1&-3&-2  \end{bmatrix} 
\xrightarrow[\text{第三行乘以 $-1$ }]{\text{第二行乘以 $1/2$ }}
\begin{bmatrix} 1&1&2&4 \\ 0&1&1&1/2 \\ 0&1&3&2  \end{bmatrix} \\ 
%&& 
\xrightarrow[\text{}]{\text{第二行乘以 $-1$ 加到第三行}}
\begin{bmatrix} 1&1&2&4 \\ 0&1&1&1/2 \\ 0&0&2&3/2  \end{bmatrix} 
\xrightarrow[\text{}]{\text{第三行乘以 $1/2$ }}
\begin{bmatrix} 1&1&2&4 \\ 0&1&1&1/2 \\ 0&0&1&3/4  \end{bmatrix} \\ 
%&& 
\xrightarrow[\text{第三行乘以 $-2$ 加到第一行}]{\text{第三行乘以 $-1$ 加到第二行}}
\begin{bmatrix} 1&1&0&5/2 \\ 0&1&0&-1/4 \\ 0&0&1&3/4  \end{bmatrix} 
\xrightarrow[\text{}]{\text{第二行乘以 $-1$ 加到第一行}}
\begin{bmatrix} 1&0&0&11/4 \\ 0&1&0&-1/4 \\ 0&0&1&3/4  \end{bmatrix}
= RREF(\overline{A}). 
\end{eqnarray*}

}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.5. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  这最后一个矩阵称为{\color{red}行最简形}，由此可以写出解
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{l}
x_1  =  11/4, \\
x_2  =  -1/4, \\
x_3  = 3/4. 
\end{array}\right.
\end{eqnarray*}
}

\item  {\color{red}注意：行初等变换的过程要用箭头（用箭头 $\rightarrow$）表示，因为每次变换都得到不同的矩阵。这与求行列式的值（用等号 $=$）是完全不同的。}

\item  {\ppr RREF} = {\ppr Reduced Row Echelon Form} = 行最简形 
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.6. 行最简形的概念}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  例子2：下述矩阵都是行最简形矩阵，

\vspace{-0.6cm}

{\footnotesize 
\begin{eqnarray*}
\begin{bmatrix} {\color{red}1}&0&*&* \\ 0&{\color{red}1}&*&* \\ 0&0&0&0  \end{bmatrix}, \hspace{0.3cm}
\begin{bmatrix} 0&{\color{red}1}&*&0&* \\ 0&0&0&{\color{red}1}&* \\ 0&0&0&0&0  \end{bmatrix}, \hspace{0.3cm}
\begin{bmatrix} {\color{red}1}&*&0&0&* \\ 0&0&{\color{red}1}&0&* \\ 0&0&0&{\color{red}1}&* \end{bmatrix}.
\end{eqnarray*}
}

\vspace{-0.3cm}

\item  {\color{red}定义：符合下述特征的矩阵称为行最简形矩阵，}
\begin{enumerate}
\item  {\color{red}行最简形矩阵是行阶梯形矩阵，即若第 $i$ 行的第一个非零元素在第$j$列，那么第 $i+1$ 行的第一个非零元素在第$j+1$列或更后面。} 
\item {\color{red}不全为零的行的第一个非零元素都是1. }
\item  {\color{red}不全为零的行的第一个非零元素1的正上方都是0. }
\item  {\color{red}全为零的行在其它行的下方。}
\end{enumerate}

\item  行最简形是矩阵通过{\color{red}行初等变换}能化到的最简形式。给定一个矩阵，它的行最简形是唯一确定的。

\end{itemize}


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.7.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize} 
\item 例子3：求线性方程组的通解，
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rrrrrr}
5x_1 &+ 3x_2 &+ 5x_3 &+ 12x_4 &=& 10, \\
2x_1 &+ 2x_2 &+ 3x_3 &+ 5x_4 &=& 4, \\
x_1 &+ 7x_2 &+ 9x_3 &+ 4x_4 &=& 2. 
\end{array}\right.
\end{eqnarray*}
}

\item 解答：用{\color{red}行初等变换}将{\color{red}增广矩阵}化为{\color{red}行最简形}，可得同解的线性方程组，
{\footnotesize 
\begin{eqnarray*}
\begin{bmatrix} 5&3&5&12&10 \\ 2&2&3&5&4 \\ 1&7&9&4&2  \end{bmatrix}
\xrightarrow[\text{}]{\text{一系列的行初等变换}}
\begin{bmatrix} 1&0&1/4&9/4&2 \\ 0&1&5/4&1/4&0 \\ 0&0&0&0&0  \end{bmatrix}. 
\end{eqnarray*}
}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.8. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  得到同解的线性方程组为 
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rrrrrr}
x_1   &            &+ (1/4)x_3 &+ (9/4)x_4 &=& 2, \\
         &+ x_2   &+ (5/4)x_3 &+ (1/4)x_4 &=& 0, \\
         &            &                 &               0 &=& 0. 
\end{array}\right.
\end{eqnarray*}
}

\item 于是将未知数 $x_3, x_4$ 移到方程右边，并忽略零方程，可得
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rrr}
x_1  &=& 2 +(-1/4)x_3 + (-9/4)x_4, \\
x_2  &=& (-5/4)x_3 + (-1/4)x_4. 
\end{array}\right.
\end{eqnarray*}
}
其中 $x_3,x_4$ 是自由未知量，可取任意实数。这是通解的一种写法。

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.9. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item 我们也可以把通解写成{\color{red}向量的线性组合}的形式，
{\footnotesize 
\begin{eqnarray*}
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4   \end{bmatrix}
=
\begin{bmatrix} 2 +(-1/4)x_3 + (-9/4)x_4 \\ (-5/4)x_3 + (-1/4)x_4 \\ x_3 \\ x_4   \end{bmatrix}
=\begin{bmatrix} 2 \\ 0 \\ 0 \\ 0   \end{bmatrix}
+x_3\begin{bmatrix} -1/4 \\ -5/4 \\ 1 \\ 0   \end{bmatrix}
+x_4\begin{bmatrix} -9/4 \\ -1/4 \\ 0 \\ 1   \end{bmatrix},
\end{eqnarray*}
}
其中 $x_3,x_4\in \mathbb{R}$, 即可取任意实数。 

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}{3.1.1. }
\begin{frame}[fragile=singleslide]{5.10.  {\ppr Python} 代码与输出结果 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{python}
import sympy as sy
A1=[5,3,5,12,10]
A2=[2,2,3,5,4]
A3=[1,7,9,4,2]
M = sy.Matrix([A1,A2,A3])
M_rref = M.rref()
s='The row echelon form and the the pivot columns:\n{}'
print(s.format(M_rref))
\end{python}

\begin{python}
The row echelon form and the the pivot columns:
(Matrix([
[1, 0, 1/4, 9/4, 2],
[0, 1, 5/4, 1/4, 0],
[0, 0,   0,   0, 0]]), (0, 1))
\end{python}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.11. 齐次线性方程组的基础解系}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}定义：非齐次线性方程组 $AX=B$ 导出的齐次线性方程组是 $AX=0$. }

\item  {\color{red}定义：齐次线性方程组 $AX=0$ 的一个基础解系，是指一些解向量组成的集合
$\{\eta_1,\cdots,\eta_s\}$, 满足如下条件：}

\begin{enumerate}
\item  {\color{red}线性方程组 $AX=0$ 的任意一个解向量 $X$ 都能写成 $\{\eta_1,\cdots,\eta_s\}$ 的线性组合的形式，即存在实数 $k_1,\cdots, k_s$ 使得 $X=k_1\eta_1+\cdots+k_s\eta_s$. }
\item  {\color{red}第1条中写法是唯一的，即对同一个解向量 $X$, 若还存在实数 $m_1,\cdots,m_s$ 使得 $X=m_1\eta_1+\cdots+m_s\eta_s$, 则一定有 $k_1=m_1, \cdots, k_s=m_s$.  }
\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.12.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  例子4：上述例子3中的非齐次线性方程组{\color{red}导出的齐次线性方程组}是
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rrrrrr}
5x_1 &+ 3x_2 &+ 5x_3 &+ 12x_4 &=& 0, \\
2x_1 &+ 2x_2 &+ 3x_3 &+ 5x_4 &=& 0, \\
x_1 &+ 7x_2 &+ 9x_3 &+ 4x_4 &=& 0. \\  
\end{array}\right.
\end{eqnarray*}
}
就是将方程右边的常数项都写成零得到的线性方程组。

\item   这个齐次线性方程组的任意一个解，都可以唯一地写成下述两个向量的线性组合的形式，这两个向量组成的集合是 $AX=0$ 的一个{\color{red}基础解系}，
{\footnotesize 
\begin{eqnarray*}
\left\{
\begin{bmatrix} -1/4 \\ -5/4 \\ 1 \\ 0   \end{bmatrix}, \,\,
\begin{bmatrix} -9/4 \\ -1/4 \\ 0 \\ 1   \end{bmatrix}
\right\}. 
\end{eqnarray*}
}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.13. 特解与通解}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item 定义：线性方程组的{\color{red}一个解}是指一个解向量 
{\footnotesize 
$$X=\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n   \end{bmatrix}. $$
}

\item 定义：线性方程组的{\color{red}解集}是指所有解向量组成的集合。
\item 定义：线性方程组的{\color{red}通解}是指解集里的任意一个解向量的一般表达式。
\item 定义：线性方程组的{\color{red}特解}是指解集里的任意一个选中的解向量。

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.14. 线性方程组的解集的结构}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}定理：设非齐次线性方程组 $AX=B$ 有特解 $\xi$, 它导出的齐次线性方程组 $AX=0$ 有基础解系 $\{\eta_1,\cdots,\eta_s\}$, 则这个非齐次线性方程组 $AX=B$ 的通解可以写成下述形式，
\begin{eqnarray*}
X=\xi + k_1\eta_1 + \cdots + k_s\eta_s, 
\end{eqnarray*}
其中 $k_1,\cdots, k_s$ 是任意实数。
}

\item  证明：

\begin{enumerate}
\item  设 $X$ 是 $AX=B$ 的任意一个解。因为 $A\xi=B$, 所以 $A(X-\xi)=0$. 
\item  因为 $\{\eta_1,\cdots,\eta_s\}$ 是 $AX=0$ 的基础解系， 所以存在实数 $k_1,\cdots,k_s$ 使得  
 \begin{eqnarray*}
X - \xi = k_1\eta_1 + \cdots + k_s\eta_s. 
\end{eqnarray*}

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.15. 课堂练习 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}

\item  用行初等变换，将增广矩阵化为行最简形，然后写出通解，
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rrrrrr}
x_1 &+ x_2 &+ 2x_3 &+ x_4 &=& 5, \\
2x_1 &+ 3x_2 &+ 2x_3 &+ 3x_4 &=& 10, \\
3x_1 &+ 2x_2 &+ 9x_3 &+ 4x_4 &=& 18.   
\end{array}\right.
\end{eqnarray*}
}

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{5.16. 课堂练习答案 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}
\item  这个非齐次线性方程组的增广矩阵的行最简形为
{\footnotesize 
\begin{eqnarray*}
%\begin{bmatrix} 1&1&2&1&5 \\ 2&3&2&3&10 \\ 3&2&9&4&18  \end{bmatrix}
%\longrightarrow
RREF(\overline{A}) = \begin{bmatrix} 1&0&0&-8&-7 \\ 0&1&0&5&6 \\ 0&0&1&2&3  \end{bmatrix}.   
\end{eqnarray*}
}
这个非齐次线性方程组的通解为 
{\footnotesize 
\begin{eqnarray*}
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4   \end{bmatrix}
=\begin{bmatrix} -7 \\ 6 \\ 3 \\ 0   \end{bmatrix}
+ x_4\begin{bmatrix} 8 \\ -5 \\ -2 \\ 1   \end{bmatrix}, \hspace{0.3cm} x_4 \in \mathbb{R}. 
\end{eqnarray*}
}

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}

